Review

This is a special post for short-form writing by Alex Mennen. Only they can create top-level comments. Comments here also appear on the Shortform Page and All Posts page.

Review

This is a special post for short-form writing by Alex Mennen. Only they can create top-level comments. Comments here also appear on the Shortform Page and All Posts page.

AlexMennen's Shortform

7Alex Mennen

Theorem: Fuzzy beliefs (as in https://www.alignmentforum.org/posts/Ajcq9xWi2fmgn8RBJ/the-credit-assignment-problem#X6fFvAHkxCPmQYB6v ) form a continuous DCPO. (At least I'm pretty sure this is true. I've only given proof sketches so far)

The relevant definitions:

A fuzzy belief over a set X is a concave function ϕ:ΔX→[0,1] such that sup(ϕ)=1 (where ΔX is the space of probability distributions on X). Fuzzy beliefs are partially ordered by ϕ≤ψ⟺∀μ∈ΔX:ϕ(μ)≥ψ(μ) . The inequalities reverse because we want to think of "more specific"/"less fuzzy" beliefs as "greater", and these are the functions with lower values; the most specific/least fuzzy beliefs are ordinary probability distributions, which are represented as the concave hull of the function assigning 1 to that probability distribution and 0 to all others; these should be the maximal fuzzy beliefs. Note that, because of the order-reversal, the supremum of a set of functions refers to their pointwise infimum.

A DCPO (directed-complete partial order) is a partial order in which every directed subset has a supremum.

In a DCPO, define x<<y to mean that for every directed set D with supD≥y, ∃d∈D such that d≥x. A DCPO is continuous if for every y , y=sup{x∣x<<y}.

Lemma: Fuzzy beliefs are a DCPO.

Proof sketch: Given a directed set D , (supD)(μ)=min{d(μ)∣d∈D} is convex, and {μ∣(supD)(μ)=1}=⋂d∈D{μ∣d(μ)=1}. Each of the sets in that intersection are non-empty, hence so are finite intersections of them since D is directed, and hence so is the whole intersection since ΔX is compact.

Lemma: ϕ<<ψ iff {μ∣ψ(μ)=1} is contained in the interior of {μ∣ϕ(μ)=1} and for every μ such that ψ(μ)≠1, ϕ(μ)>ψ(μ).

Proof sketch: If supD≥ψ, then ⋂d∈D{μ∣d(μ)=1}⊆{μ∣ψ(μ)=1} , so by compactness of ΔX and directedness of D, there should be d∈D such that {μ∣d(μ)=1}⊆int({μ∣ϕ(μ)=1}). Similarly, for each μ such that ψ(μ)≠1, there should be dμ∈D such that dμ(μ)<ϕ(μ) . By compactness, there should be some finite subset of {d}∪{dμ∣ψ(μ)≠1} such that any upper bound for all of them is at least ϕ .

Lemma: ψ=sup{ϕ∣{μ∣ψ(μ)=1}⊆int{μ∣ϕ(μ)=1},∀μψ(μ)≠1→ϕ(μ)>ψ(μ)} .

Proof: clear?